Please help with all, I’ll mark brainliest as long as you tell me how you got that answer.
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ANSWER
EXPLANATION
For each of the three events, there are 8 possible outcomes, so the probability of each event is,
• Event A:, there must be only odd numbers in the last two rolls, which are either ,EOO or OOO,. This event has 2 favorable outcomes, ,so its probability is 1/4,.
,• Event B,: this is the complementary event to event A. Here the last two rolls must be even numbers - both or only one of the last two: ,OOE, EEO, EOE, OEE, EEE, OEO,. This event has 6 favorable outcomes, so ,its probability is 3/4,.
,• Event C,: this event is when the first and last rolls are even numbers in the same event. This is either ,EOE or EEE,. Since this event has 2 favorable outcomes, ,its probability is 1/4,.
Point F is on line a, so it does represent Josiah's distance at a certain time. Also, point F is below line b, so it represents a distance that is less than Chana's distance. This is a distance-time graph problem.
<h3>What is the proof for the above?</h3>
Recall that Josiah had a head start of 10 meters and he skates at 2 meters per second.
Since Y is the function that represents the distance in meters from the finished line, by observation, it is clear to see that all the factors that are related to his race are adequately represented in:
y = 10 + 2x
Where 10 is the head start in meters
2 is the rate at which he skates per second; and
x is the unknown amount of time in seconds.
Given that the point F sits over 25 seconds,
that is F(y) = 10 + 2 * 25
= 60 meters.
Hence, Point F is on line a, so it does represent Josiah's distance at exactly 25 seconds.
Learn more about distance-time graphs at:
brainly.com/question/4931057
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